F73ZF2/3ZH3: Survival Models II and III

June 2006 Examination Solutions

Section A

1. (a) We have λ(t, z) = λ0(t) exp(β′z). (1)

(b) Non-smoker, under 60. (1)

(c) The ratio is exp(β1 + β2) = exp(0.3) = 1.35. (1)

[Total 3]

2. (a) tpwwx is the probability that a life in state w at age x remains in state w for a further

time t. tpwwx is the probability that a life in state w at age x is in state w at age x+ t;

in particular, tpwwx says nothing about where the life is between ages x and x+ t. tp

wwx

is the larger. (3)

(b) We have

t+dtpwwx = tp

wwx (1− α dt+ o(dt)).

Rearranging, dividing by dt and letting dt→ 0 we get (with y(t) = tpwwx )

d

dty(t) = −α y(t).

The solution to this equation is y(t) = tpwwx = e−αt. (3)

(c) We have

t+dtpwwx = tp

wwx (1− α dt+ o(dt)) + tp

wfx (β dt+ o(dt)).

Rearranging, dividing by dt and letting dt → 0 we get (with tpwwx = z(t) and tp

wfx =

1− tpwwx = 1− z(t))

d

dtz(t) = −α z(t) + β (1− z(t)) = β − (α + β) z(t).

(3)

[Total 9]

3. (a) We use the standard formula

3p70 = exp

{

−∫ 3

0

µ70+t dt

}

= exp

{

−(∫ 1

0

0.04 dt+

∫ 2

1

0.045 dt+

∫ 3

2

0.05 dt

)}

= e−(0.04+0.045+0.05) = e−0.135 = 0.874

and so 3q70 = 0.126. (3)

1

(b) The probability that two such lives survive and one dies is

3 3p270 3q70 = 3× 0.8742 × 0.126 = 0.289.

(2)

[Total 5]

4. (a) µx+ 1

2

= dv. (1)

(b) and (c) µx+ 1

2

= qx/(1− 12qx). (1)(1)

(d) Hence equating (a), (b) and (c) we find

d

v= µx+ 1

2

=qx

(1− 12qx)

and solving for qx gives the actuarial estimate

qx =d

v + 12d.

(2)

[Total 5]

5. There are claims at times 1, 2, 7, 8, 11, 15, 16, 20 and 23 so the partial likelihood is

L(β) ∝1

8 + 7eβ·

eβ

7 + 7eβ·

1

5 + 4eβ·

eβ

4 + 4eβ·

1

4 + 3eβ·

1

2 + 2eβ·

eβ

1 + 2eβ·

1

1 + 1eβ

∝e3β

(1 + eβ)4(8 + 7eβ)(5 + 4eβ)(4 + 3eβ)(1 + 2eβ)

with partial log likelihood

`(β) = 3β − 4 log(1 + eβ)− log(8 + 7eβ)− log(5 + 4eβ)− log(4 + 3eβ)− log(1 + 2eβ).

⇒d`

dβ= 3−

4eβ

1 + eβ−

7eβ

8 + 7eβ−

4eβ

5 + 4eβ−

3eβ

4 + 3eβ−

2eβ

1 + 2eβ.

(5)

(b) Use Figure 2. The maximum partial likelihood estimate of β is −0.525 approxi-

mately. For the standard error we need need the slope of d`/dβ which is

(0.1 + 0.2)/(−0.58 + 0.42) = −0.3/0.16 = −1.875.

Thus the SE is√

1/1.875 =√

(0.533) = 0.73 approximately.

For the z-test we compute z = −0.525/0.73 = −0.72 so there is little evidence againstH0. (5)

(c) For the likelihood ratio test we use Figure 1. `(β) = −10.45 while `(0) = −10.7.Hence

−2 log Λ = −2(−10.7 + 10.45) = 0.5

2

and since −2 log Λ ∼ χ21 under H0 there is no worthwhile evidence against the null

hypothesis. (3)

(d) We estimated β as −0.525 and so the ratio of the hazard for female to male is

exp(−0.525) = 0.6, very different from 1. However, the standard error of β is very

large relative to the estimated value and the value of the LRT is only 0.5, a very

small value on the scale of χ21. No definite conclusion can be drawn on any differences

between the two groups. (2)

[Total 15]

6. (a)(i) The values of b−aqx+a under UDD are: q (lives 1 to 5), 0.5q/(1− 0.2q) (lives 6 to

8) and 0.5q/(1− 0.5q) (lives 9 and 10). Hence the likelihood is

L(q) ∝ q2 × (1− q)3 × (0.5q/(1− 0.2q))2 × (1− 0.7q)/(1− 0.2q)× ((1− q)/(1− 0.5q))2

∝ q4(1− q)5(1− 0.7q)/[(1− 0.2q)3(1− 0.5q)2]

∝ q4(1− q)5(10− 7q)/[(5− q)3(2− q)2]

⇒ `1(q) = 4 log q + 5 log(1− q) + log(10− 7q)− 3 log(5− q)− 2 log(2− q).

(ii) From the plot q ≈ 0.47. (6)

(b)(i) We require the waiting time

v = 0.3 + 0.5 + 3 ∗ 1 + 0.2 + 0.4 + 0.5 + 2 ∗ 0.5 = 5.9

⇒ `2(µ) = d log µ− vµ = 4 log µ− 5.9µ

since there are d = 4 deaths. The MLE of µ is µ = d/v = 4/5.9 = 0.678.

(ii) Under the constant force of mortality assumption q = 1− e−µ and

so µ = − log(1− q). Substituting in `2(µ) from (b)(i) we find

`3(q) = 4 log(− log(1− q)) + 5.9 log(1− q)

as required. Differentiating wrt q we find

d`3(q)

dq= 4×

1

− log(1− q)×

1

1− q−

5.9

1− q= 0⇒ log(1− q) = −

4

5.9

and so q3 = 1− exp(−4/5.9) = 0.492.

(iii) We know that q = 1 − e−µ so from µ = 0.678 we find q = 1 − e−0.678 = 0.492, as

expected. (8)

[Total 14]

3

7. (a) We have

t+dtpwux = tp

wux (1− ρ dt− ν dt+ o(dt)) + tp

wwx (σ dt+ o(dt)).

Rearranging, dividing by dt and letting dt→ 0 we get

∂

∂ttpwux = −(ρ+ ν) tp

wux + σ tp

wwx .

(3)

(b) Use symmetry with tpwux . So

∂

∂ttpuwx = −(σ + µ) tp

uwx + ρ tp

uux .

(1)

(c) Suppose the transfers occur in small intervals of time δt1, δt2 and δt3 at March 1,

July 1 and October 1 respectively. The probability of the given sequence of events is

e−2(σ+µ) [σ δt1 + o(δt1)] e−4(ρ+ν) [ρ δt2 + o(δt2)] e

−3(σ+µ) [µ δt3 + o(δt3)].

Dividing through by δti and letting δti → 0, i = 1, 2, 3, gives the contribution as

L(α, ρ, µ, ν) ∝ e−5σσ1 × e−5µµ1 × e−4ρρ1 × e−4νν0.

(4)

(d) Standard formulae give

σ =TwuWw

=200

4000; µ =

TwdWw

=8

4000

so σ = 0.05 and µ = 0.002. Standard errors are

SE(σ) =

√

σ

Ww

=

√

0.05

4000≈ 0.0035; SE(µ) =

√

µ

Ww

=

√

0.002

4000≈ 0.0007.

(4)

[Total 12]

4

8. (a) and (b) The plot is given below and strongly suggests the fitted model is fine.

60 62 64 66 68 70 72

−5.

0−

4.8

−4.

6−

4.4

−4.

2−

4.0

−3.

8

Age

log(

mu)

(5)

(c) We have log(µx) = −12.145 + 0.1172x so at x = 60 we have d60 = 52250 ∗exp(−12.145 + 0.1172 ∗ 60) ≈ 314.5.

The standardized residual is z = (d− d)/√

d = (312− 321)/√321 = −0.50.

The plot of the residuals is

60 62 64 66 68 70 72

−1.

5−

1.0

−0.

50.

00.

51.

01.

5

Age

Res

idua

ls, z

.x

No large residuals or suspicious patterns. Graduation fine. (6)

(d) We have Φ(0.75) = 0.67 so so the numbers of residuals in the four cells are: 4 less

than −0.67, 2 between −0.67 and 0, 4 between 0 and 0.67, and 3 greater than 0.67.

The expected number in each cell is 13/4 = 3.25. Hence

χ2 = [(4− 3.25)2 + (2− 3.25)2 + (4− 3.25)2 + (3− 3.25)2]/3.25 = 2.75/3.25

which is less than 1 and so no worthwhile evidence of lack of fit. (4)

5

(e) The weight should be inversely proportional to the variance of the dependent vari-

able, log µx. Using the ∆-method we find

Var(log µx) = Var(log dx) ≈ (1/dx)2Var(dx) ≈ (1/dx)

2 ∗ dx = 1/dx

so wx = dx is a good choice. (4)

[Total 19]

9. (a) Let PCx,t and PD

x,t be the number of lives aged x at time t according to the census

and death definitions of age respectively. Then

PDx,t ≈

1

2

(

PCx−1,t + PC

x,t

)

(assuming a uniform distribution of birthdays).

Now, assuming that PDx,t is linear between census dates, we have (using the trapezium

rule)

Ecx ≈

1

2

1∑

t=0

(

PDx,t + PD

x,t+1

)

which gives

Ec51 ≈

1

2

(

PD51,0 + 2PD

51,1 + PD51,2

)

≈1

4

(

PC50,0 + PC

51,0 + 2PC50,1 + 2PC

51,1 + PC50,2 + PC

51,2

)

=1

4(5451 + 6002 + 2× 4515 + 2× 5534 + 5934 + 4428)

=41913

4= 10478.25.

Then,

µ51 =d51

Ec51

=76

10478.25= 0.00725

and

q51 = 1− e−µ51 = 0.00723.

Since the age definition for deaths is age x next birthday, the estimates refer to ages

from 50 to 51. Therefore, µ50 applies to exact age 50.5 and q50 applies to exact age 50.

(8)

(b) (i) Using the actuarial estimate of initial exposed to risk we have

q51 =d51

E51

≈d51

Ec51 +

12d51

=76

10478.25 + 762

=76

10516.25= 0.00723.

6

(ii) It is preferred to use the method in (a) to avoid assumptions involved in approxi-

mating the initial exposed to risk Ex. In addition, (a) is based on the ‘multiple-state’

model, which is more general (than the binomial model) and has good statistical prop-

erties.

(4)

(c) Under both the Poisson and the multiple-state model we have

var(µx) =µxEcx

≈µxEcx

⇒ se(µx) =

√

0.00751

10478.25= 0.00085.

The two estimates are consistent. (3)

(d) (i) Calendar year rate interval ranging from 1 January to 31 December.

(ii) At the start of the rate interval, exact ages range from x− 1 to x. Thus, assuming

uniformly distributed birthdays over the calendar year, the average age at the start of

the rate interval is x− 0.5. It follows that the estimates of µ51 and q51 apply to exact

ages 51 and 50.5 respectively.

(3)

[Total 18]

[Grand total: 100]

7